Does $E(X_n)\rightarrow E(X)$ as $n\to\infty$

Question

If there is a sequence of random variables ${X_n}$ converging almost surely to $X$, therm is it true that $E(X_n)\rightarrow E(X)$ as $n\to\infty$ ? Only thing given is that $E(X_n)\le 23$ for all $n$.

I am not getting how to do it. I can't use DCT here, can I?

Possible duplicate of Does almost sure convergence implies convergence of the mean? — leonbloy, Nov 04 '16 at 14:34

Gono · Accepted Answer · 2016-11-04T14:32:41.177

2

No it's not.

Consider: $$X_n(\omega) = \begin{cases} n & \mbox{ if } \omega\in [0,\frac{1}{n}] \\ 0 & \mbox{ otherwise} \end{cases}$$

Then $X_n \to X$ where $X \equiv 0$, $E[X_n] = 1 \le 23$, so $\lim\limits_{n\to\infty} E[X_n] = 1$ but $E[X] = 0$

edited Nov 04 '16 at 14:32

answered Nov 04 '16 at 13:52

Gono

5,788

What is $x$... ? – leonbloy Nov 04 '16 at 14:23
X is constant 0 – Gono Nov 04 '16 at 14:26
I meant (and I wrote) lowercase $x$ - in the "if". – leonbloy Nov 04 '16 at 14:28
I guess now it's clear :-) – Gono Nov 04 '16 at 14:33

Does $E(X_n)\rightarrow E(X)$ as $n\to\infty$

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